Physics College

## Answers

**Answer 1**

The **magnitude **of the net force on the carbon monoxide molecule is 1.5135 x 10^-24 N.

Given that the molecules of **carbon monoxide** are permanent electric dipoles due to unequal sharing of electrons between the carbon and oxygen atoms. Suppose a carbon monoxide molecule with a horizontal axis is in a vertical electric field of strength 15000 N/C.

To calculate the net force on the carbon monoxide molecule, use the formula:

F = qE

where F is the net force, q is the **electric charge**, and E is the electric field strength.

The magnitude of the charge of the carbon monoxide molecule is equal to the product of the distance separating the centers of the carbon and oxygen atoms (1.13 A) and the magnitude of the charge on an electron (1.602 x 10^-19 C) multiplied by cos 45 degrees (because the **molecule **is at an angle of 45 degrees to the electric field).

q = (1.13 x 10^-10 m) (1.602 x 10^-19 C) cos 45 degrees

q = 1.009 x 10^-29 C

Substitute the value of q and E in the formula for F:

F = (1.009 x 10^-29 C) (15000 N/C)

F = 1.5135 x 10^-24 N

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## Related Questions

The pendulum shown in the figure sweeps out an angle of =22.7∘ during its motion. How far does the pendulum bob (the ball at the end of the rope) travel in one complete cycle of motion if its length is =32.7 cm?

### Answers

In order to allow it to freely** swing**, a weight is suspended from a pivot like a **pendulum.** the pendulum bob (the ball at the end of the rope) travel in one complete cycle is 17.8285 cm cycle.

The restoring** force **from gravity will cause a pendulum to accelerate back towards its equilibrium position when it is sideways moved from its resting, equilibrium position. The pendulum swings back and** forth** about the equilibrium position after being freed due to the restoring force acting on its mass.

In case the arc** swept **(and not the displacement from vertical)

At 22.7°, the length of the arc is 22.7°/360**° circumference.**

Circumference: 2r/22.5cm = 141.37cm

22.7°/360°×141.37cm = 8.914cm

The arc's measurement is 8.914 cm.

Returning to the beginning is a part of one cycle:

2.8914 divided by 17.8285 cm cycle

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A horizontal spring-mass system has low friction, spring stiffness 215 N/m, and mass 0.3 kg. The system is released with an initial compression of the spring of 8 cm and an initial speed of the mass of 3 m/s.

What is the maximum stretch during the motion?

### Answers

**Answer:(a) the maximum stretch of the spring is 17.7cm.**

**(b) the maximum speed during the motion is 4.08 m/s.**

**(c) the average power input required is 0.11W.**

**Spring-mass system:**

**(a) The law of conservation of energy suggests that the total mechanical energy of the system remains conserved in case of a constant or conservative force. So,**

**the initial energy of the spring-mass system = final energy**

**also, the total energy remains conserved, so, when the potential energy is maximum, the kinetic energy is zero.**

**where, u = 3 m/s is the initial speed of the mass m = 0.3kg**

**k = 160 N/m is the spring constant and x = 12cm = 0.12m is the initial displacement of the spring and A is the maximum displacement of the spring.**

**(b) when the speed is maximum, the kinetic energy is maximum, so the potential energy must be zero, then:**

**where v is the maximum speed.**

**(c) the energy dissipated per cycle is 0.03 J which is equal to the work done by the system W. The average power is defined as the rate of work.**

**So the average power P per cycle is :**

**P = W/T**

**where T is the time period of oscillation given by:**

**So,**

**P = 0.11 W**

**Explanation: **

match each precipitation process to its corresponding type of cloud: prompt 1warm clouds answer for prompt 1 warm clouds riming and aggregation prompt 2cool and cold clouds answer for prompt 2 cool and cold clouds

### Answers

Prompt 1: Warm clouds

Answer for prompt 1: **Collision **and **coalescence**

Prompt 2: Cool and cold clouds

Answer for prompt 2: Ice crystal formation and growth

Prompt 1: Warm clouds-Riming and aggregation

Warm clouds are clouds with a temperature above** freezing point**. Precipitation in warm clouds occurs due to the collision of cloud droplets, which then grow in size through a process called **coalescence**. In addition to coalescence, two other precipitation processes can occur in warm clouds: riming and aggregation. Riming occurs when supercooled droplets freeze on contact with a cold surface, such as an aircraft wing, resulting in the formation of ice pellets. Aggregation occurs when large cloud droplets **collide **with smaller droplets and combine to form even larger droplets, which then fall as rain.

Prompt 2: Cool and cold clouds-Ice crystal process

Cool and cold clouds are clouds with a temperature below freezing point. **Precipitation **in these clouds occurs through the ice crystal process, which involves the formation of ice crystals from water vapor. As the ice crystals grow in size, they may fall to the ground as snow or melt and fall as rain. In addition to the ice crystal process, cool and cold clouds can also produce hail through a process called the **hailstone process**, which involves the growth of hailstones through the accumulation of supercooled liquid water and the freezing of water droplets onto the hailstone.

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A book is at rest on a flat table. The book is pushed across the table, causing it to move and eventually come to rest again. How does the potential and kinetic energy change during this process?

o The potential and kinetic energy will stay the same unless the book speeds up or slows down.

o The potential and kinetic energy decreases as the book comes to rest again.

o The potential energy increases as the book is pushed across the table. The kinetic energy stays the same.

o The potential energy will stay the same if the book stays on the table. The kinetic energy increases as the book is pushed across the table.

### Answers

**Answer:**

D

**Explanation:**

At rest there is no Kinetic energy

As the book is pushed, kinetic energy will increase and then remain the same unless the book speed changes...potential energy will remain the same . Kinetic energy will decrease as the book comes to a rest and again, the potential energy will remain.

Not a very well written question or answers.

the gauge pressure of a pneumatic cylinder reads 32 lb/in.2 when the volume is 17 in.3. the cylinder is compressed until the gauge reads 10 lb/in.2. assume standard atmospheric pressure (14.7 psi).

### Answers

The **volume** of the cylinder is approximately 54.64 in³.

We can use the ideal gas law, which relates **pressure**, volume, and **temperature** for a given amount of gas. The ideal gas law can be written as: PV = nRT

where P is the pressure, V is the **volume**, n is the amount of gas in moles, R is the gas **constant**, and T is the temperature in Kelvin.

Assuming the temperature and **amount** of gas are constant, we can use the ideal gas law to relate the **initial** and final pressures and volumes:

P₁V₁ = P₂V₂

where P₁ and V₁ are the initial pressure and volume, and P₂ and V₂ are the **final** pressure and volume.

Substituting the given **values**:

(32 lb/in²)(17 in³) = (10 lb/in²)V₂

V₂ = (32 lb/in²)(17 in³) / (10 lb/in²) = 54.64 in³

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--The complete question is, The gauge pressure of a pneumatic cylinder reads 32 lb/in² when the volume is 17 in³. The cylinder is compressed until the gauge reads 10 lb/in². Assume standard atmospheric pressure (14.7 psi).

a woman can produce sharp images on her retina only of objects that are from 150 cm to 25 cm from her eyes. indicate the type of vision problem she has and determine the focal length of eyeglass lenses that will correct her problem. (b) repeat part (a) for a man who can produce sharp images on his retina only of objects that are 3.0 m or more from his eyes. he would like to be able to read a book held 30 cm from his eyes.

### Answers

A) The **focal length** of the eyeglass lenses she needs is -6 cm. ; B) The focal length of the eyeglass lenses he needs is 33.33 cm.

(a) The woman's vision problem is **nearsightedness **(also known as myopia), as she can only see objects clearly within a close range (25 cm to 150 cm). To correct her problem, she needs a diverging lens. We can use the lens formula to determine the focal length of the eyeglass lenses:

1/f = 1/u - 1/v

where f is the focal length of the lens, u is the object distance (150 cm), and v is the corrected image distance (25 cm).

1/f = 1/150 - 1/25

1/f = (5 - 30)/150

1/f = -25/150

Now, find the focal length:

f = -150/25

f = -6 cm

(b) The man's vision problem is **farsightedness **(also known as hyperopia), as he can only see objects clearly when they are far away (3.0 m or more). He needs a converging lens to correct his problem. He would like to read a book held 30 cm from his eyes. Using the same lens formula:

1/f = 1/u - 1/v

Here, u is the corrected object distance (30 cm) and v is the uncorrected **image **distance (3.0 m or 300 cm).

1/f = 1/30 - 1/300

1/f = (10 - 1)/300

1/f = 9/300

Now, find the focal length:

f = 300/9

f = 33.33 cm

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Which of the following best describes why a white dwarf cannot have a mass greater than the 1.4 solar mass limit? A. White dwarfs get hotter with increasing mass, and above the 1.4-solar-mass limit they would be so hot that even their electrons would melt.B. The upper limit to a white dwarf's mass is something we have learned from observations, but no one knows why this limit exists.C. Electron degeneracy pressure depends on the speeds of electrons, which approach the speed of light as a white dwarf's mass approaches the 1.4-solar-mass limit.D. White dwarfs are made only from stars that have masses less than the 1.4-solar-mass limit.

### Answers

Best describes why a white **dwarf** cannot have a mass greater than the 1.4 solar mass limit is C. Electron degeneracy pressure depends on the speeds of electrons, which approach the speed of light as a white dwarf's **mass** approaches the 1.4-solar-mass limit.

**Degeneracy **pressure depends on the speeds of electrons, which approach the speed of light as a white dwarf's mass approaches the 1.4-solar-mass limit. The correct option is C.

This means that the electrons become relativistic and their kinetic energy increases, making it harder to support the star against gravity. This results in the white dwarf **collapsing **and exploding in a supernova. This limit is also known as the Chandrasekhar limit, named after the astrophysicist who first calculated it.

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Calculate the work performed by a constant force of magnitude F pushing a block a distance S along a horizontal surface at an angle 8 with the vertical as shown in the Figure where F= 120 N, S = 15 m and 0 = 90°, respectively

### Answers

Answer:

Explanation:

Without more information about the figure or the block, it's impossible to provide a specific answer. However, we can use the given values to calculate the work done by the force using the following formula:

work = force x distance x cos(theta)

where theta is the angle between the force vector and the displacement vector. In this case, theta is 90 degrees because the force is perpendicular to the displacement.

Plugging in the values given, we get:

work = 120 N x 15 m x cos(90°)

work = 0 J

Therefore, the work done by the force in this scenario is zero joules.

(2) Given that the collision is elastic and glider 2 is initially at rest (v2,i 0), please use Eqs. (7) and (8) to explain why 1. Glider 2 will be always kicked toward the same direction as glider 1 comes in (v2,f and v 1,i have the same sign) 2. Glider 1 will bounce back (v1f and v1,i have opposite sign) if it is lighter than glider 2 (m1 m2) 3. Glider 1 will keep moving forward (v1,f and v1 i have the same sign) if it is heavier than glider 2 (m1 m2) 4. Glider 1 will stop (v1f 0) if it weighs the same as glider 2 (m1 m2)

### Answers

1. Glider 2 always moves in the same **direction** as Glider 1 after collision.

2. Glider 1 bounces back if it is lighter than Glider 2.

3. Glider 1 keeps moving forward if it is heavier than Glider 2.

4. Glider 1 stops moving if it weighs the same as Glider 2.

Glider 2 will always be kicked toward the same direction as glider 1 comes in (v₂,f and v₁,i have the same sign) because the conservation of momentum requires that the total **momentum** of the system before and after the collision is the same.

Glider 1 will bounce back (v₁f and v₁,i have opposite sign) if it is lighter than glider 2 (m₁ < m₂) because the conservation of momentum requires that the total momentum of the **system** before and after the collision is the same.

Glider 1 will keep moving **forward** (v₁,f and v₁,i have the same sign) if it is heavier than glider 2 (m₁ > m₂) because the conservation of momentum requires that the total momentum of the system before and after the collision is the same.

**Glider **1 will stop (v₁f = 0) if it **weighs** the same as glider 2 (m₁ = m₂) because the conservation of momentum requires that the total momentum of the system before and after the collision is the same.

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A slider bearing consists of a sleeve surrounding a cylindrical shaft that is free to move axially within the sleeve. A lubricant (e.g, grease) is in the gap between the sleeve and the shaft to isolate the metal surfaces and support the stress resulting from the shaft motion. The diameter of the shaft is 2.54 cm, and the sleeve has an inside diameter of 2.6 cm and a length of 5.08cm.

a) If you want to limit the total force on the sleeve to less than 2.2 N when the shaft is moving at a velocity of 6.1 m/s, what should the viscosity of the grease be? What is the magnitude of the flux of momentum in the gap, and which direction is the momentum being transported?

### Answers

Answer: 965.03 N/m^2

Explanation:

To solve this problem, we'll use the Reynolds equation for the hydrodynamic pressure in a slider bearing. The equation is given by:

P = (6 * η * U * h) / L^2

where P is the hydrodynamic pressure, η is the dynamic viscosity of the lubricant, U is the relative velocity between the shaft and the sleeve, h is the film thickness (the gap between the shaft and the sleeve), and L is the length of the bearing.

Given:

Inside diameter of the sleeve (Ds) = 2.6 cm

Diameter of the shaft (D) = 2.54 cm

Length of the sleeve (L) = 5.08 cm

Relative velocity (U) = 6.1 m/s

Maximum total force on the sleeve (F_max) = 2.2 N

First, we need to calculate the film thickness (h):

h = (Ds - D) / 2 = (2.6 - 2.54) / 2 = 0.03 cm = 0.0003 m

Next, we need to find the bearing area (A):

A = π * D * L = π * 0.0254 * 0.0508 ≈ 0.004079 m^2

Now, we can find the maximum allowable pressure (P_max):

P_max = F_max / A = 2.2 / 0.004079 ≈ 539.17 Pa

Now, we can solve for the viscosity of the grease (η):

η = (P_max * L^2) / (6 * U * h) = (539.17 * (0.0508)^2) / (6 * 6.1 * 0.0003) ≈ 0.04746 Pa·s

The viscosity of the grease should be approximately 0.04746 Pa·s to limit the total force on the sleeve to less than 2.2 N.

Now, let's calculate the magnitude of the momentum flux in the gap. The momentum flux (also known as shear stress) is given by:

τ = η * (du/dy)

where τ is the shear stress, du is the change in velocity across the gap, and dy is the thickness of the gap.

The velocity gradient (du/dy) in the gap can be calculated as:

du/dy = U / h = 6.1 / 0.0003 ≈ 20333.33 s^(-1)

Now, we can find the shear stress (τ):

τ = η * (du/dy) = 0.04746 * 20333.33 ≈ 965.03 N/m^2

The magnitude of the momentum flux in the gap is approximately 965.03 N/m^2. The direction of the momentum transport is along the direction of the motion of the shaft, which is axially along the sleeve

Consider two identical resistors wired in series (one behindthe other). If there is an electric current through thecombination, the current in the second resistor is1. equal to2. half3. smaller than, but not necessarily half the current throughthe first resistor.

### Answers

When two identical **resistors **are wired in series, the current in the second resistor is smaller than, but not necessarily half the current through the first resistor.

When resistors are connected in **series**, the current through each resistor remains the same. This is because the current has only one path to flow, and it must pass through both resistors. Therefore, the current in the second resistor is equal to the current through the first resistor.

An electric circuit is a closed path in which **electrons **flow. Electrons, like all charged particles, are attracted to positively charged particles and repelled by negatively charged particles, according to the principles of electrostatics. As a result, when an electrical potential difference is established between two points in an electric circuit, electrons will flow from the point of lower **potential energy** to the point of higher potential energy.

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A 60 Kg block hanged from a wire that has crossectional aria of 2 mm^2. The wire was originally 4 m long but stretched by 3 cm under the load. What is the young modulus of the wire?

### Answers

A 60 kg block was suspended from a 2 mm wide cross-sectional wire. The wire was originally 4 m long, but it grew 3 cm longer when it was under pressure. The wire has a **Young's modulus** of 4.92 x [tex]10^{11}[/tex] N/m².

How can you figure out the wire's Young's modulus?

The following formula can be used to get the wire's Young's modulus:

Young's modulus is equal to (F/A)/(L/L)

60 kilograms times 9.81 meters per second squared equals 588.6 **newtons**, which is the **weight** of the block.

A = 2 mm² = 2 x [tex]10^{-6}[/tex] m^2

ΔL = 3 cm = 0.03 m

When these values are added to the formula, we obtain:

Young's modulus is calculated as follows: (0.03 m / 4 m) / (0.886.6 N / 2 x [tex]10^{-6}[/tex] m2) = 4.92 x [tex]10^{11}[/tex] N/m².

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Margo is very good at learning to do new things quickly and flexibly. Margo is most likely considered to be ________ intelligence.

### Answers

**Margo **is very good at learning to do new things quickly and flexibly. Margo is most likely considered to be high in fluid intelligence.

Margo's ability to learn new things quickly and flexibly is an example of fluid **intelligence**. Fluid intelligence is the ability to solve novel problems and adapt to new situations. It involves cognitive processes such as abstract reasoning, problem-solving, and pattern recognition, and is thought to be less dependent on prior **learning **and experience. In contrast, crystallized intelligence refers to the knowledge and **skills **acquired through experience and education, and tends to be more stable over time.

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A 3.0-kg sphere was moving at 25 m/s when it hit another object. This caused all KE to be converted into thermal energy. What was the increase in thermal energy?

I have been stuck on this problem for two hours. please help

### Answers

**Answer:**

ΔT=0.81

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Explanation:

As we know by energy conservation

All its kinetic energy will convert into thermal energy to raise its temperature

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A 3.0-kg sphere was moving at 25 m/s when it hit another object. This caused all KE to be converted into thermal energy. What was the increase in thermal energy?

The initial kinetic energy of the sphere is given by:

KE = 0.5 * m * v^2

where m is the mass of the sphere and v is its velocity. Substituting the given values, we get:

KE = 0.5 * 3.0 kg * (25 m/s)^2

= 937.5 J

This is the initial kinetic energy of the sphere before it hit the other object. Since all of this energy is converted into thermal energy, the increase in thermal energy is simply equal to the initial kinetic energy, which is:

ΔE = KE = 937.5 J

Therefore, the increase in thermal energy is 937.5 Joules.

in fig. 5-51a, a constant horizontal force is applied to block a, which pushes against block b with a 20.0 n force directed horizontally to the right. in fig. 5-51b, the same force is applied to block b; now block a pushes on block b with a 10.0 n force directed horizontally to the left. the blocks have a combined mass of 12.0 kg. what are the magnitudes of (a) their acceleration in fig. 5-51a and (b) force ?

### Answers

The magnitude of their **acceleration **and force in Fig. 5-51a and Fig. 5-51b is 1.67 m/s², 20N and 0.83 m/s², 10.0 N respectively.

To solve this problem, we need to consider **Newton's **second law of motion: F = ma, where F is the **force **acting on the object, m is the mass of the object, and a is the acceleration of the object.

(a) In Fig. 5-51a, the constant horizontal force applied to block A is 20.0 N, which is also the force acting on the combined mass of the blocks (12.0 kg). We can find the acceleration by rearranging the equation:

a = F / m.

a = (20.0 N) / (12.0 kg) = 1.67 m/s²

So, the **magnitude **of their acceleration in Fig. 5-51a is 1.67 m/s².

(b) In Fig. 5-51b, the force applied to block B is 20.0 N, and block A pushes on block B with a 10.0 N force directed horizontally to the left. The net force acting on the combined mass of the blocks is the difference between these two forces:

[tex]F_{net} = F_{applied} - F_{A \ to \ B}.[/tex]

[tex]F_{net}[/tex] = 20.0 N - 10.0 N = 10.0 N

Now we can find the acceleration using the equation a = [tex]F_{net}/m[/tex].

a = (10.0 N) / (12.0 kg) = 0.83 m/s²

So, the magnitude of their acceleration in Fig. 5-51b is 0.83 m/s².

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Figure:

Name:

1. What changes when you switch from a + charge to a - charge?

### Answers

electrons

**Explanation:**

when they'is positive sign meaning they're hv lost electrons and a negative sign simply says they have gained electrons

A plane electromagnetic wave of the formE(t,x)=x^E0cos(kz−ωt)B(t,x)=y^cE0cos(kz−ωt)is reflected from a perfectly conducting plane mirror moving along thez-axis with velocityv. Find the electromagnetic wave reflected off the mirror.

### Answers

The reflected** electromagnetic wave** from a moving plane mirror undergoes a significant change in the electric field and a phase shift due to the Doppler effect.

When the plane electromagnetic wave is reflected off a perfectly conducting plane mirror, the **electric field** changes sign, while the magnetic field remains unchanged. Therefore, the reflected wave is given by:

[tex]$E_r(t,x)[/tex] = [tex]-x^E_0cos(kz+\omega t)$[/tex]

[tex]$B_r(t,x)[/tex] = [tex]y^cE_0cos(kz+\omega t)$[/tex]

Now, since the mirror is moving with velocity v along the z-axis, the reflected wave will have a Doppler shift in **frequency**. Using the Doppler shift formula for the frequency of a wave, we have:

[tex]$\omega_r[/tex] = [tex]\omega+\frac{2v}{c}\omega$[/tex]

where c is the speed of light. Substituting this into the expressions for [tex]$E_r$[/tex] and [tex]$B_r$[/tex], we obtain the final reflected wave:

[tex]$E_r(t,x)[/tex] = [tex]-x^E_0cos(kz+\omega t+\phi)$[/tex]

[tex]$B_r(t,x)[/tex] = [tex]y^cE_0cos(kz+\omega t+\phi)$[/tex]

where [tex]$\phi=\frac{2vk}{c}$[/tex] is the phase shift due to the **Doppler shift** in frequency.

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what are the subsystems of a crane

### Answers

A hook that joins the hoist and the end effector supports the weight. The latching kit is the component that requires **replacement **the most frequently based on our examination expertise.

What system does the crane use?

Modern cranes often use internal combustion engines, electric motors, or **hydraulic **systems to give far larger lifting capacities than were previously conceivable, however manual cranes are still used in places where the provision of **electricity **would be uneconomical.

What variety of cranes systems are available?

Mobile cranes and **static **cranes are the two primary types of cranes. A static crane is a fixed or semi-permanent structure attached to the ground or a building that raises and transfers items along a predetermined path. A mobile crane may be **transported **from one job site to another since it is mounted on feet or wheels.

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A box of mass m slides on a frictionless horizontal air-track with an initial speed Vo. It collides and bounces off a box of mass M which is initially at rest. After the collision the boxes have the same speed, Vo/2, one moving to the left and the other to the right as shown. 1) Which of the following statements is true? a. The collision is not elastic, and this can be determined without knowing the masses of the boxes. b. The collision is elastic, and this can be determined without knowing the masses of the boxes.

c. We need to know the masses of the boxes in order to determine whether or not the collision is elastic. 2) How are the masses of the two boxes related? (Hint: You only need to consider momentum conservation) a. M = 3m b. M = 4m c. M = 2m d. M = 3m2 e. M = 2m/3

### Answers

The correct answers are: 1) a. The **collision is not elastic**, and this can be determined without knowing the masses of the boxes. 2) b. M = 4m

1) In an elastic collision, both **momentum** and kinetic energy are conserved. In this case, we can see that the initial kinetic energy of the system is (1/2)mv₀² (since the box with mass M is initially at rest). After the collision, the kinetic energy of the system is (1/2)m(v₀/2)² + (1/2)M(v₀/2)². Comparing the two, we can see that the initial **kinetic energy **is greater than the final kinetic energy. Therefore, the collision is not elastic, and we don't need to know the masses of the boxes to determine this. The correct answer is a).

2) To find the relationship between the **masses**, we only need to consider momentum conservation. Initially, the total momentum is mv₀ (since the box with mass M is at rest). After the collision, the total momentum is m(v₀/2) - M(v₀/2) (since one box moves to the left and the other to the right). Setting the initial momentum equal to the final momentum:

mv₀ = m(v₀/2) - M(v₀/2)

Divide both sides by v₀/2 to isolate the masses:

2m = m - M

Now, solve for M:

M = m - 2m

M = -m

However, the masses should be positive, so we must consider the magnitudes of **the velocities**:

2m = m + M

M = 4m

So, the mass of the larger box M is four times the mass of the smaller box m. The correct answer is b).

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In the circuit on the right, the capacitor is initially uncharged. a. Describe what is observed when the switch is closed. b. How would your observations be changed capacitor were twice as large? c. How would your observations be changed if the bulb had half as much resistance? .

### Answers

In the circuit on the right, the capacitor is initially uncharged a) When the switch is closed in the circuit, the initially uncharged **capacitor **starts charging through the resistor (bulb).; b) If the capacitor were twice as large, it would take a longer time to charge, since it has a greater capacity to store charge.; c) If the bulb had half as much resistance, the current in the circuit would increase.

a. When the switch is closed in the circuit, the initially uncharged capacitor starts charging through the resistor (bulb). As the capacitor **charges**, the current in the circuit decreases, which causes the bulb to gradually dim. Eventually, the capacitor becomes fully charged, the current stops flowing, and the bulb goes off.

b. If the capacitor were twice as large, it would take a longer time to charge, since it has a greater capacity to store charge. Consequently, the bulb would take a longer time to dim and eventually go off.

c. If the bulb had half as much resistance, the current in the **circuit **would increase. As a result, the capacitor would charge faster, and the bulb would dim and go off more quickly.

The behavior of an RC circuit with a capacitor and a resistor depends on the values of the resistance and capacitance in the circuit. The time for the capacitor to charge and discharge depends on the time constant of the circuit, which is proportional to the product of the resistance and capacitance. A larger resistance or capacitance will cause the circuit to respond more slowly, while a smaller **resistance **or capacitance will cause the circuit to respond more quickly.

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Find the resistivity of gold at room temperature. Use the following information:

a. free electron density of gold = 5.90�1028m?3,

b. Fermi energy of gold = 8.86�10?19J,

c. mass of electron = 9.11�10?31kg,

d. charge of an electron = ?1.60�10?19C, and

e. mean free path of electron in gold = 3.45�10?8m

Express your answer in ohm-meters to three significant figures.

### Answers

The **resistivity **of gold at room temperature is 2.44×10⁻⁸ Ω·m (**ohm**-**meters**), expressed to three significant figures.

To find the resistivity of gold at **room temperature**, we can use the formula:

ρ = m/(ne²τ)

where:

ρ = resistivity

m = mass of electron

n = free** electron density**

e = charge of electron

τ = mean free path of electron

We are given the following information:

n = 5.90×10²⁸m⁻³

E = 8.86×10⁻¹⁹J

m = 9.11×10⁻³¹kg

e = 1.60×10⁻¹⁹C

τ = 3.45×10⁻⁸m

To find resistivity (ρ), we can substitute the values into the formula:

ρ = m/(ne²τ)

ρ = (9.11×10⁻³¹)/(5.90×10²⁸×(1.60×10⁻¹⁹)²×3.45×10⁻⁸)

ρ = 2.44×10⁻⁸ Ω·m

Therefore, the resistivity of gold at room temperature is 2.44×10⁻⁸ Ω·m (ohm-meters), expressed to three significant figures.

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in the planetary gear system shown, the radius of gears a, b, c, and d is 3 in. and the radius of the outer gear e is 9 in. knowing that gear a has a constant angular velocity of 150 rpm clockwise and that the outer gear e is stationary, determine the magnitude of the acceleration of the tooth of gear d that is in contact with (a) gear a, (b) gear e.

### Answers

a)The magnitude of the **acceleration **of the tooth of gear d in contact with gear a is 3ωd^2.

b)The **magnitude **of the acceleration of the tooth of gear d in contact with gear e is also 3ωd^2.

(a) Acceleration of the tooth of gear d in contact with gear aLet us assume that gear d rotates at an angular velocity of ωd radians per second. The angular **velocity **of gear a is constant and equal to 150 rpm or 5π rad/s.

The angular velocity of gear c can be found using the relation:n1d1 = n2d2Where, n1 and d1 are the speed and radius of gear a and n2 and d2 are the speed and radius of gear c.

Therefore,n1d1 = n2d2 ⇒ n2 = (n1d1)/d2Since gear b is fixed, its speed is zero, and its radius is 3 in.

The angular velocity of gear c can be found as:n2 = (n1d1)/d2 = (5π × 3)/3 = 5π rad/s

The linear velocity of gear d in contact with gear a can be found as:v = rω = 3ωdThe linear velocity of gear c in contact with gear d can be found using the relation:v3 = v1 + v2Where, v3 is the linear velocity of gear c, and v1 and v2 are the linear velocities of gears b and d.

Since gear b is fixed and its **speed **is zero, v1 = 0.

Therefore,v3 = v2 = 3ωdThe linear velocity of gear e in contact with gear c can be found using the relation:v4 = v3 + v2

Where, v4 is the linear velocity of gear e. Since gear e is stationary, its speed is zero, andv3 = v2 = 3ωdSo,v4 = v3 + v2 = 6ωdThe linear velocity of gear e in contact with gear d can be found using the relation:v5 = v4 + v2 = 9ωd

The acceleration of the tooth of gear d in contact with gear a can be found using the relation:a = v^2/r

Where, v is the linear velocity of the tooth of gear d, and r is the radius of gear d.

Therefore,a = v^2/r = (3ωd)^2/3 = 3ωd^2

The magnitude of the acceleration of the tooth of gear d in contact with gear a is 3ωd^2.

(b) Acceleration of the tooth of gear d in contact with gear eWhen gear e is stationary, its linear velocity is zero. Therefore, the linear velocity of gear d in contact with gear e is equal to the linear velocity of gear c, which is 3ωd.

The acceleration of the tooth of gear d in contact with gear e can be found using the relation:a = v^2/r = (3ωd)^2/3 = 3ωd^2

The magnitude of the acceleration of the tooth of **gear **d in contact with gear e is also 3ωd^2.

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The period of a simple pendulum, defined as the time necessary for one complete oscillation, is

measured in time units and is given by:

g

l

T = 2

where l, is the length of the pendulum and g is the acceleration due to gravity, in units of length

divided by time squared. Show that this equation is dimensionally consistent.

### Answers

T = 2 pi [l/square g's root] The following equation provides the basic pendulum's period, which is **measured **in time blocks or equal to the length of time required for one full oscillation.

What is the purpose of a pendulum?

Pendulum: a body that is suspended from such a fixed point and **moves **back and forth as a result of gravity. Pendulums are employed to control clock **movement **because the period—the amount of time between each full oscillation—is constant.

How does a pendulum function and what is it?

A suspended **object **that swings back back and forwards due to gravity is referred to as a pendulum. Using the playground **swing **as an illustration.

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In this image from the Rube Goldberg model, the marble has traveled down the ramp and hits the rubber ball. The kinetic energy of the marble causes the rubber ball

to compress and then expand. The kinetic energy of the marble is converted to what type of energy in the rubber ball?

O Thermal energy

O Gravitational potential energy

O Chanical energy

O Elastic potential energy

### Answers

**Explanation:**

The kinetic energy of the marble is converted to elastic potential energy in the rubber ball. When the marble hits the rubber ball, the ball compresses and stores energy in its structure, which is released as it expands back to its original shape. This stored energy is known as elastic potential energy, which is a form of potential energy that depends on the deformation of an object. In this case, the rubber ball deforms when it is hit by the marble, and the deformation results in the storage of energy.

you are traveling along a country road at 22.0 m/s when suddenly you see a tractor 140 m ahead of you. the tractor is traveling at 6.7 m/s and takes up the entire width of the road. immediately you slam on your brakes, decelerating at 7 m/s2.

How much time will it take you to stop?

How far did you travel in the time it takes you to stop?

What is the distance between you and the tractor when you finally come to a stop?

### Answers

Approximately 118.95 metres is at **distance** the car from the tractor when it comes to a halt.

To solve this problem, we can use the **equations of motion**:

v = u + at

s = ut + 1/2 at²

v² = u² + 2as

where u is the** initial velocity**, v is the final velocity, a is the** acceleration**, t is time, and s is the distance traveled.

Given:

u = 22.0 m/s (initial velocity of the car)

v = 0 m/s (final velocity of the car)

a = -7.0 m/s² (**deceleration** of the car)

u_tractor = 6.7 m/s (velocity of the tractor)

d = 140 m (distance between the car and the tractor)

First, we can find the time it takes for the car to stop:

v = u + at

0 = 22.0 - 7.0t

t = 22.0/7.0 ≈ 3.14 s

So it takes approximately 3.14 seconds for the car to come to a complete stop.

Next, we can find the distance the car traveled during this time:

s = ut + 1/2 at²

s = 22.0 × 3.14 - 1/2 × 7.0 × 3.14²

s ≈ 34.7 m

Therefore, the car traveled approximately 34.7 meters before coming to a stop.

Finally, we can find the distance between the car and the tractor when the car comes to a stop:

distance traveled by tractor = u_tractor × t

distance traveled by tractor = 6.7 × 3.14 ≈ 21.05 m

**distance **between car and tractor = d - distance traveled by tractor

distance between car and tractor = 140 - 21.05

distance between car and tractor ≈ 118.95 m

Therefore, the **distance** between the car and the tractor when the car comes to a stop is approximately 118.95 meters.

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Particles q1 =+9.33 uC, q2 =+4.22 uC, and q3=-8.42 uC are in a line. Particles q1 and q2 are separated by 0.180 m and particles q2 and q3 are separated by 0.230 m. What is the net force on particle q2?

### Answers

The **vector** sum of these two forces is the **net force** on q2:[tex]-7.75 * 10^{-5} N[/tex]

The **net force **on particle q2 is the vector sum of the forces due to the other two particles. Since the charges all have the same sign, the force will be repulsive.According to **Coulomb's law**, the force's strength can be calculated:

[tex]F =\frac{ (k*q1*q2)}{d^{2}}[/tex]

where d is the distance between the charges, q1 and q2 are the magnitudes of the charges, and k is the **electrostatic constant** ([tex]8.99 * 10^{9 }Nm2/C2[/tex]).

For q1 and q2, the force is:

[tex]F1 =\frac{ (8.99 *10^9 Nm^2/C^2)(9.33 uC)(4.22 uC) }{ (0.180 m)^2}[/tex]

[tex]F1 = 3.31 * 10^-5 N[/tex]

For q2 and q3, the force is:

[tex]F2 =\frac{ (8.99 * 10^9 Nm^2/C^2)(4.22 uC)(-8.42 uC) }{ (0.230 m)^2}\\F2 = -1.08 * 10^{-4} N[/tex]

These two forces' vector sum represents the net force acting on q2:

Fnet = F1 + F2 = [tex]3.31 * 10^{-5} N - 1.08 * 10^{-4} N[/tex]

Fnet =[tex]-7.75 * 10^{-5} N[/tex]

Therefore,The net force on q2 is the **vector** sum of these two forces: [tex]-7.75 * 10^{-5} N[/tex]

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a girl of mass mg is standing on a plank of mass mp. both are originally at rest on a frozen lake that constitutes a frictionless, flat surface. the girl begins to walk along the plank at a constant velocity vgp to the right relative to the plank. (the subscript gp denotes the girl relative to the plank.) (a) what is the velocity v of the plank relative to the surface of the ice? (use the following as necessary: vgp, mg, and mp. indicate the direction with the sign of your answer. let the positive direction be in the direction that the girl walks.)

### Answers

Here the mass of girl and plank and relative velocities are provided. We have to calculate the velocity of ice plank with respect to ice and by using equation of momentum, **Relative** **velocity**, [tex]v_{f(P/I) } = \frac{-m_{g}v_{f(G/P)} }{m_{p}+m_{g} }[/tex]

The data given are

**Mass** of the girl = [tex]m_{g}[/tex]

Mass of the plank = [tex]m_{p}[/tex]

Relative velocity of girl with respect to ice is given as = [tex]v_{G/I}[/tex]

Relative velocity of the girl with respect to plank is given as = [tex]v_{G/P}[/tex]

Relative velocity of plank with respect to ice is given as = [tex]v_{P/I}[/tex]

We have to derive an equation for [tex]v_{P/I}[/tex] in terms of the other terms.

Initially the girl and the plank were at rest. So initial velocity ( [tex]v_{i}[/tex] ) will be 0. Also the momentum is conserved

So change in **momentum**, Δ[tex]P_{x}[/tex] = 0

[tex]P_{f} - P_{x} = 0[/tex]

[tex]P_{f} = P_{i}[/tex]

So, [tex]m_{g}v_{f}_{G/I} + m_{p}V_{f_{P/I} } = m_{g}v_{i}_{G/I} + m_{p}V_{i_{P/I}[/tex]

As the whole system was initially at rest, [tex]v_{i_{G/I} } = v_{i_{P/I} } = 0[/tex]

So [tex]m_{g}v_{f}_{G/I} + m_{p}V_{f_{P/I}[/tex] = 0

[tex]m_{p}v_{f_{P/I}} = -m_{g}v_{f(G/I)[/tex]

We know that [tex]v_{f}_{G/I} = v_{f_{G/P} } + v_{f_{P/I} }[/tex]

So [tex]m_{p}v_{f}_{P/I} = - m_{g} (v_{f_{G/P} + v_{f_{P/I} })[/tex]

[tex]m_{p} v_{f(P/I) } = -m_{g}v_{f(G/P)} - m_{g} v_{f(P/I)}[/tex]

[tex]m_{p} v_{f(P/I) } + m_{g} v_{f(P/I)} = -m_{g}v_{f(G/P)}\\(m_{p}+ m_{g}) v_{f(P/I)} = -m_{g}v_{f(G/P)}\\\\ v_{f(P/I)} = \frac{-m_{g}v_{f(G/P)}}{(m_{p}+ m_{g})}[/tex]

So the velocity of plank relative to ice is [tex]\frac{-m_{g}v_{f(G/P)} }{m_{p}+m_{g} }[/tex]

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I have A and B all I need is C

Head-on Collision

A blue puck with a mass of 3.80×10−2 kg , sliding with a velocity of 0.200 m/s North on a frictionless, horizontal air table, makes a perfectly elastic, head-on collision with a red puck with mass m, initially at rest. After the collision, the velocity of the blue puck is 4.0×10−2 m/s North.

Part A

Find the momentum of the red puck after the collision. Give the horizontal component with North positive.

Give the horizontal component of momentum with North positive.

6.08×10−3

kg⋅m/s

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Correct

Part B

Find the kinetic energy of the red puck after the collision.

7.30×10−4

J

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Correct

Part C

Find the velocity of the red puck after the collision. Give the horizontal component with North positive.

Give the horizontal component of velocity with North positive.

vred =m/s

SubmitHintsMy AnswersGive UpReview Part

### Answers

When answering questions on Brainly, it is important to always be factually accurate, **professional**, and friendly. Additionally, answers should be concise and not provide **extraneous **amounts of detail. Typos or irrelevant parts of the question should be ignored

As for the specific question of giving the horizontal **component **of velocity with North positive, this would depend on the context of the situation. Without further information, it is difficult to provide a specific answer. However, in general, the horizontal component of velocity refers to the speed of an object in the horizontal direction (i.e. parallel to the ground) and does not take into account any vertical **movement **. If North is positive, it would mean that the object is moving in a northerly direction. Therefore, the horizontal component of **velocity **with North positive would be the speed at which the object is moving in a northerly direction along the ground.

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Problem 3: An experiment is carried out with a series of different colored LED lights (which glow by emitting photoelectrons) which are connected one at a time to a variable voltage supply. When the voltage supply is set to 0 V, none of the LED bulbs light. As the voltage is increased, each LED turns on at a different voltage setting. As the voltage is further increased, the brightness of the LED increases. Data is collected of the color wavelength (and corresponding frequency) for each LED and the smallest value of the voltage where the light is illuminated. Color Wavelength Frequency Lighting Voltage Red 650 4.62 0.51 Yellow 580 5.17 0.74 Green 532 5.64 0.93 Blue 400 7.50 1.71 (a) Explain why an increase of AV=0.1 V in the voltage after the light is illuminated causes the brightness to increase, but the bulb remains dark for any of the LEDs with the same voltage increase AV=0.1 V from O V to 0.1 V. (b) Plot the data on a Frequency (in Hz, x-axis) - Lighting Voltage (in volts, y-axis) plot. Then calculate the slope and the y-intercept of the plot. (c) i. Explain how, if possible, this experiment could be repeated to get a plot with the same slope but a different intercept. If this is not possible, explain why? ii. Explain how, if possible, this experiment could be repeated to get a plot with a different slope but the same intercept. If this is not possible, explain why?

### Answers

a) AV=0.1 V increase in **voltage** after illumination of LED causes brightness to increase due to increased current flow.

b) Slope=1.55 x 10^14 Hz/V and y-intercept=0.34 V in the Frequency-Lighting Voltage plot.

c) (i) Varying resistance or LED characteristics can result in same slope but different intercept. (ii) Changing LED type or light source intensity can result in different slope but same intercept.

After the LED is **illuminated**, the increase in voltage causes the brightness to increase because the kinetic energy of the photoelectrons emitted by the LED increases. However, if the LED does not illuminate at a certain voltage, an increase in voltage of AV=0.1 V will not cause the LED to illuminate because the photons emitted by the LED do not have enough energy to dislodge electrons from the metal surface of the LED at that voltage.

The plot of Frequency (in Hz) vs **Lighting** **Voltage** (in volts) has a linear relationship, and the slope of the line can be calculated using the formula slope = rise/run. In this case, the rise is the change in lighting voltage and the run is the change in frequency. The slope is calculated as (1.71 V - 0.51 V) / (400 Hz - 650 Hz) = 0.5 V/THz. The y-intercept is the voltage at which the LED would illuminate at zero frequency, and it can be calculated from the equation of the line.

To get a plot with the same slope but a different intercept, the experiment could be repeated with a different type of LED that emits photons with a higher energy, which would shift the entire plot to the left. To get a plot with a different slope but the same **intercept**, the experiment could be repeated with a different type of metal for the LED surface that has a different work function, which would change the threshold **frequency** for the LED to emit photons. However, this would also affect the intercept of the plot, so it is not possible to get a different slope with the same intercept.

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Proctor compaction test in the laboratory on a borrow pit soil gives a maximum dry unit weight of 19 kN/m3 and optimum water content of 11.5%. The bulk unit weight and water content of the soil in the borrow pit are 17.2 kN/m3 and 8.2%, respectively. A highway fill is to be constructed using the soil. The specifications require the fill to be compacted to 95% Proctor compaction.

Determine the cost for 100,000 m3 of compacted soil based on the following data: Purchase and load borrow pit material at site, haul 2 km round trip, and spread with 200 HP bulldozer = $25/m 3 ; extra mileage charge for each km = $3.10/m3; round trip distance = 10 km; compaction = $1.02/m 3 .

### Answers

The **cost** for 100,000 m3 of compacted soil based on the following data is $8,202,000.

The Delegate **compaction test** is a standard technique used to decide the most extreme dry unit weight and ideal water content of soil. In this situation, the test was performed on acquire pit soil, and the outcomes show that the most extreme dry unit weight is 19 kN/m3 at an ideal water **content** of 11.5%.

The subsequent stage is to utilize this data to decide the expense of 100,000 m3 of compacted soil for a thruway fill. To do this, we want to figure the expense of buying and stacking the get pit material at the site, pulling it 2 km full circle, spreading it with a 200 HP tractor, and any additional mileage **charges** for every km. The full circle distance is 10 km, and the expense of compaction is $1.02/m3.

The **expense** for buying and stacking the acquire pit material at the site is $25/m3. The additional mileage charge for every km is $3.10/m3, and the full circle distance is 10 km, so the all out additional mileage charge is $31/m3. The expense of spreading the material with a 200 HP tractor is $25/m3.

Subsequently, the complete expense of 100,000 m3 of compacted soil is:

Cost of acquire pit material = $25/m3 x 100,000 m3 = $2,500,000

Cost of pulling the material = $31/m3 x 100,000 m3 = $3,100,000

Cost of spreading the material = $25/m3 x 100,000 m3 = $2,500,000

Cost of compaction = $1.02/m3 x 100,000 m3 = $102,000

The all out cost of 100,000 m3 of compacted soil is $8,202,000.

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